Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 5s (2025) 481 https://internationalpubls.com Principal Intersection Graph of Commutative Rings Lamine Ngom1, Mankagna Albert Diompy 2 1,2Department of Mathematics and Computer Science, University of Cheikh Anta Diop, Dakar, Senegal Email Id: laminengo89@hotmail.fr1 Corresponding author: albertdiompy@yahoo.fr Article History: Received: 13-10-2024 Revised: 28-11-2024 Accepted: 09-12-2024 Abstract: Let 𝑅 be a commutative ring. The principal intersection graph of a commutative ring 𝑅, noted 𝐺𝑐(𝑅), consist of all proper ideals of 𝑅 as vertices. Two distinct vertices 𝐼 and 𝐽 are adjacent if 𝐼 ∩ 𝐽 β‰  0 and either 𝐼 or 𝐽 is a principal (cyclic) ideal. In this paper, we investigate some properties from graph theory of 𝐺𝑐(𝑅) and its algebraic properties where 𝑅 is a ring. Keywords: Principal intersection graph, Principal ideal domain, Ore domain, Bezout domain, Connected graph, Complete graph, Hamiltonian graph. AMS subject Classification: 13A70, 05C25, 05C45. 1. Introduction The intersection graph of ideals of a ring R is the graph having the set of all ideals as its set of vertices. Two distinct vertices I and J are adjacent if and only if their intersection is non-zero idea and either I or J is a principal (cyclic) ideal. Intersection graph were introduced by Bosak in 1964 [6]. Since, particular intersection graph like small intersection graph, prime intersection graph, semisimple intersection graph are studied respectively in [3, 1, 11, 5, 7]. Recently, several properties of these kinds of graphs were investigated by many authors as Ansari-Toroghy, Nikmehr - Soleymanzadeh and Alwan in 2016; 2017 and 2023 respectively. In this paper, R is a commutative ring with identity (or eventually a domain). Here, we introduce a particular intersection graph Gc(R) named Principal Intersection Graph, whose set of vertices is the proper ideals of R. We will study the algebraic properties of Gc(R) and also its properties when seen as a graph. This paper is organized as follow: in the first section, we recall some properties of rings and graph theory. In the second section, we study connectedness, completeness, k-partite and Hamiltonian properties of this intersection graph. We gave a characterization of the connectedness, completeness and Hamiltonian properties of Gc(R) as a principal ideal domain, an Ore domain and a Bezout domain. 2. Definitions and preliminary results Definition 2.1.: Definitions from ring theory β€’ An ideal I of commutative ring R is principal written as I = aR for some a ∈ R, if it is generated by one element. β€’ A ring R is principal if every proper ideal is a principal ideal. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 5s (2025) 482 https://internationalpubls.com β€’ A ring R is an Ore ring if it satisfies the Ore Condition. That is: For all elements a and b in R, aR ∩ bR β‰  {0}. β€’ A ring R is a domain if it has no zero-divisor. β€’ A ring R is an Ore ring if for all elements a and b in R, aR ∩ bR β‰  {0}. β€’ A principal ideal domain is a domain such that every ideal of R is a principal ideal. β€’ Bezout ring is a domain in which for any to elements a, b ∈ R, there is n β‰₯ 0 such that Ran + Rb n is a principal ideal. Definition 2.2.: Definitions from graph theory β€’ Let I and J two distinct vertices, Iβˆ’ J means that I and J are adjacent. β€’ The degree of a vertex I of graph Gc(R)) which denoted by deg(I) is the number of edges incident on I. β€’ If |V(Gc(R)))| > 2, a path from I to J is a sequence of adjacent vertices Iβˆ’ I1 βˆ’ I2 βˆ’β‹―βˆ’ In βˆ’ J, where Ii ∈ V(Gc(R). β€’ The length a path graph of a graph is the number of edges in this path. β€’ A path using k distinct vertices has length kβˆ’ 1. β€’ The distance between two distinct vertices I and J is denoted by d(I; J) is the length of the shortest path connecting I and J. β€’ If there is not a path between I and J, d(I; J) = 0. β€’ The number of vertices of Gc(R) is the order of the graph. β€’ The diameter of a graph Gc(R)) is diam(Gc(R))) = sup{d(I; J)/I; J ∈ V(Gc(R)))}. β€’ A graph Gc(R)) is connected, if for any vertices I and J of Gc(R)) there is a path between I and J. If not, Gc(R)) is disconnected. β€’ A closed path I βˆ’ I1 βˆ’ I2 βˆ’β‹―βˆ’ In βˆ’ I is a cycle. β€’ The girth of Gc(R)) is the length of the shortest cycle in Gc(R)). β€’ A Hamiltonian cycle is a cycle that contains every vertex of the graph. β€’ A Hamiltonian graph is graph containing a Hamiltonian cycle. β€’ A graph with no loop or multiple edges is a simple graph. Proposition 2.3. Let R be a ring. Gc(R) is an empty graph if and only if R is a field. Proof: β€’ β‡’) It is obvious that if Gc(R) is empty graph then Gc(R) has no vertices. Since vertices of Gc(R) are the proper ideals of R, then R has proper ideal, that is R is a field. β€’ ⇐) Conversely, if R is a field, R has no proper ideal. β–« Lemma 2.4. If the graph Gc(R) is a null graph, then for all (a; b) ∈ R βˆ– {1} Γ— R βˆ– {1}, ab = 0. Proof: Assume that Gc(R) is a null graph. Let a β‰  1 and b β‰  1 be two elements of R. Since Gc(R) is a null graph, then aR ∩ bR = {0}. Therefore, we have that, ab ∈ aR ∩ bR. Thus, ab = 0 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 5s (2025) 483 https://internationalpubls.com Proposition 2.5. Let R be a commutative nonzero ring. The graph Gc(R) is a null graph if and only if R = {0; 1}. Proof: β€’ ⇐) It is clear that if R = {0; 1}, then Gc(R) is a null graph. β€’ β‡’) Let 1 β‰  x ∈ R and Gc(R) a null graph. Take y ∈ R such that y β‰  1 and 1βˆ’ y β‰  1, then xy = 0. By lemma 2.4, x = x βˆ’ 0 = xβˆ’ xy = x(1βˆ’ y) = 0. Since R is commutative nonzero ring, then R = {0; 1}. Example 2.6. If p is prime integer, the graph Gc(β„€p) is null graph. Lemma 2.7. If R is a domain, then Gc(R) is a connected. Proof: Let I and J to vertices of Gc(R) . Since I and J are proper ideals of R, there exists a β‰  0 and b β‰  0 in I and J, respectively, such that ab β‰  0. So, we have ab ∈ I ∩ J implies that I ∩ J β‰  0. If one of the ideals I or J is principal, then I and J are adjacent. Moreover, I and J are not principal ideals. Since I ∩ J β‰  0, let 0 β‰  c ∈ I ∩ J and put K = cR. Then I ∩ K β‰  0 and K ∩ J β‰  0. Thus I βˆ’ Kβˆ’ J is a path between I and J. 3. Connectedness, Completeness, Hamiltonian graph Lemma 3.1. If R is a domain, every connected graph Gc(R) is complete. Theorem 3.2. Let R be a domain. The followings statements are equivalents: 1. Gc(R) is a connected graph; 2. Gc(R) is a complete graph; 3. R is an Ore domain. Proof: 1. β‡’(2) Follows from Lemma 3.1 2. β‡’(3) Let a and b to non-zero elements in R. Put on I = aR and J = bR. Since Gp(R is connected, I and J principal ideals, then I ∩ J β‰  0. Hence aR ∩ bR β‰  0 and ab β‰  0. That is R is an Ore domain. 3. β‡’ (1) follows from Lemma 2.7 Proposition 3.3. Let R be domain. If Gc(R) is a connected graph, then diam(Gc(R)) ≀ 2. Proof: Let I and J be to vertices of Gc(R). β€’ If I ∩ J β‰  0, such that at least of of them is principal, then I βˆ’ J. Thus d(I, J) = 1. β€’ If I ∩ J β‰  0, I and J both none principal, there are non-zero elements a and b such that Iβˆ’ cR βˆ’ J with c = ab. Thus d(I, J) = 2. β€’ If I ∩ J = 0 for all nonzero a ∈ I and b ∈ J, aR ∩ bR = 0 which contradicts Theorem 3.2. Hence, diam(Gc(R)) ≀ 2 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 5s (2025) 484 https://internationalpubls.com Proposition 3.4. Let R be a ring. The following statements are equivalents. 1. Gc(R) is a complete graph; 2. R is essential and R has at most one non-principal ideal. Proof: β€’ (1)β‡’ (2). Assume that Gc(R) is complete and let I be a proper ideal of R. By definition, the vertex I is adjacent to any others vertex, that is I is essential ideal. Then R is essential ring. Assume again that R has at least two proper ideals which are not principal. Let I1 and I2 be two non- principal ideals of R. The vertices I1 and I2 can not be adjacent; that is Gc(R) is not complete. Then R has at most one non-principal ideal. β€’ (2)β‡’ (1). Let J and K be two vertices of Gc(R). Since R is essential, then J ∩ K β‰  0. Since R has at most one non-principal ideal, we have two possible cases: either J and K are principal, or exactly one between J and K is principal – If J and K are principal, J and K adjacent vertices. – If one between J and K is principal, J and K are adjacent vertices. β–« The result follows. Lemma 3.5. The graph Gc(R) of a principal ideal domain R is a complete graph. Proof: Let I and J two proper ideals of R. Since I and J nonzero ideals, there is non-zero elements a and b in R such that a ∈ I and b ∈ J. Hence, 0 β‰  ab ∈ aR ∩ bR βŠ† I ∩ J implies that I and J are adjacent. Example 3.6. The graph Gc(β„€) is complete because for all n, m ∈ β„€, nβ„€ ∩mβ„€ β‰  0. Corollary 3.7. If R a field, then Gc(R[x]) is a complete graph. Lemma 3.8. The graph Gc(R) of an kΓΆthe ring R is a complete graph. Proof: Let I and J be two proper ideals of R. Since R is a kΓΆthe ring, there is non-zero elements a and b in R such that a ∈ I and b ∈ J. Hence aR ∩ bR βŠ‚ I ∩ J implies that I and J are adjacent. Lemma 3.9. 1. Gc(R) is a complete graph if and only if R is an essential domain which has at most one non- principal ideal. 2. If R has more than one non-principal ideal, then Gc(R) is a disconnected graph. Proof: 1. If Gc(R) is a complete graph, by proposition 3.4, it has at most one non-principal ideal. For all a and b two non-zero elements in R, aR ∩ bR = abR. Since Gc(R) is a complete, then ab β‰  0. Conversely, if R is an essential domain which has at most one non-principal ideal, then Gc(R) is complete by proposition 3.4 2. It is clear that two non-principal ideals of R can not be adjacent vertices of the graph Gc(R). β–« Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 5s (2025) 485 https://internationalpubls.com Theorem 3.10. Let R be a Bezout ring. The followings statements are equivalents: 1. Gc(R) is a complete graph; 2. R is a principal ideal domain. Proof: 1. β‡’(2). Since Gc(R) is a complete graph, there is at most one non-principal ideal. Let I this ideal. For all a1 ∈ I, I1 = Ra1 is adjacent I, that is I1 ∩ I β‰  0. If I = I1, I is principal. Otherwise, there exist a2 ∈ I βˆ– I1 and let I2 = Ra1 + Ra2. If I β‰  I2 = Ra1 + Ra2, there exists a3 ∈ I βˆ– I2 and let I3 = Ra1 + a2 + Ra3. Inductively, let In = Ra1 +β‹―+ Ran. If I β‰  In, we choose an+1 ∈ I βˆ– In. Since Gc(R) is complete, the chain I1 βˆ’ I2 βˆ’β‹―βˆ’ In must be finite. Moreover, the ideal In = Ra1 +β‹―+ Ran is principal because R is Bezout domain. This is a contradiction. Then R is a principal ideal domain. 2. β‡’(1) follows from Lemma 3.5 Corollary 3.11. Let R be a Bezout domain. The followings statements are equivalents: 1. Gc(R) is a complete graph; 2. R is a principal ideal domain. 3. For two ideals I and J, I ∩ J = 0 implies I = 0 or J = 0. 4. For all (a; b) ∈ R2, aR ∩ bR = 0 implies a = 0 or b = 0. 5. Every non-zero ideal of R is indecomposable. Remark 3.12. 1. If R is Bezout domain, for all vertices I1, I2, β‹― , In ∈ Gc(R), I1 βˆ’ I2 βˆ’β‹―βˆ’ In βˆ’ I1 is a cycle. 2. girth(Gc(R)) = 3 Proposition 3.13. If R is Bezout domain, N and K two vertices of Gc(R) such that K βŠ‚ N, then deg(K) ≀ deg(N). Proof: Let N and K two vertices of Gc(R) such that K βŠ‚ N. If J is another vertex of Gc(R) then J ∩ K β‰  0. Since R is Bezout principal ideal domain and J ∩ K βŠ‚ J ∩ J ∩ N, then J ∩ N β‰  0. Theorem 3.14. The followings statements are equivalents in a Bezout domain R. 1. Gc(R) is a complete; 2. R is an integral domain and has at most one non-principal ideal; 3. R is a principal ideal domain. Proof: β€’ (1)⇔ (2) follows from Lemma 3.9 β€’ (3)⇔ (1) follows from Theorem 3.10 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 5s (2025) 486 https://internationalpubls.com Corollary 3.15. The graph Gc(R) of a Bezout domain is a regular graph. Proof: Since R is a Bezout domain, Gc(R) is complete in view of Theorem 3.14; Then Gc(R) is regular graph. Proposition 3.16. If R is an Ore domain with k proper principal ideals, then the clique number w(Gc(R)) = k. Proof: Let I1, I2, …, Ik the k proper principal ideals, Ik+1, Ik+2, … , In the nβˆ’ k proper non-principal ideals of R. Since R is an Ore domain, for every vertex Ii for i ∈ {1,2, … , k} Ii ∩ Ij β‰  0 for j > k. That is Ii for i ∈ {1,2, … , k} adjacent to each other vertex in the graph. Then the graph induced by the path {Ik+1, Ik+2, … , In} is complete. Thus w(Gc(R)) = k. Here we recall a result from [5] Theorem 3.17. Let R a domain with k proper non-principal ideals and kβ€² proper principal ideals of R. The followings statements are equivalents. 1. Gc(R) is a simple graph and kβ€² β‰₯ k; 2. Gc(R) is Hamiltonian graph. Proof: (1) β‡’ (2). The order of Gc(R) is n = kβ€²+ k. Let Lk the set of non-principal ideals, Lkβ€² the set of principal ideals, (I, J) a pair of non-adjacent vertices of Gc(R). Three cases are possibles: – case 1: If I and J are non - principal ideals, then deg(I) + deg(J) = 2kβ€² β‰₯ kβ€²+ k = n. – case 2: If I and J are principal ideals, then deg(I) + deg(J) = 2(nβˆ’ 1) β‰₯ kβ€²+ k = n. – case 3: If exactly one ideal between I or J is principal, deg(I) + deg(J) = kβ€²+ 2(nβˆ’ 1) β‰₯ n. By Ore Theorem, Gc(R) is a Hamiltonian graph. (2) β‡’ (1). Assume that Gc(R) is not a simple graph or kβ€² < k; – If Gc(R) is not a simple graph clearly Gc(R) is not Hamiltonian graph. – If kβ€² < k, since n = k+ kβ€² there is no cycle containing every vertex of Gc(R). That is, there is no Hamiltonian cycle. Then Gc(R) is not Hamiltonian. Acknowledgements The authors would like to express their sincere thanks for the referee for his/her helpful suggestions and comments. 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